/**
 * 这道题需要注意精度，因为$10^{-9} \le b_i \le 10^9$，所以$r$最大会达到$2\times
 *10^9$，`int`装不下 题解给了一种$O(n)$的双指针做法
 **/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
ll a[N], b[N];
int n, m;
ll L[N], R[N];
bool check(ll r) {
  for (int i = 0; i < m; ++i) {
    L[i] = b[i] - r;
    R[i] = b[i] + r;
  }
  int j = 0;
  for (int i = 0; i < n; ++i) {
    ll cur = a[i];
    bool ok = false;
    while (L[j] <= cur) {
      if (R[j] >= cur) {
        ok = true;
        break;
      } else {
        j++;
        if (j > m) {
          return false;
        }
      }
    }
    if (!ok) {
      return false;
    }
  }
  return true;
}

int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  ll lo = 0, hi = 2e9 + 7;
  long long res = -1;
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    // cout << mid << ' ';
    if (check(mid)) {
      res = mid;
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  cout << res << endl;
}
